A child flies a kite at a height of 90 ft , the wind carrying the kite horizontally away from the child at a rate of \( 30 \mathrm{ft} / \mathrm{sec} \). How fast must the child let out the string when the kite is 150 ft away from the child? The child must let out the string at a rate of \( \square \mathrm{ft} / \mathrm{sec} \) when the kite is 150 ft away from the child. (Simplify your answer.)

To solve this problem, we can use the Pythagorean theorem. Let \( x \) be the horizontal distance from the child to the kite, \( y \) be the height of the kite (90 ft), and \( s \) be the length of the string. According to the Pythagorean theorem: \[ s^2 = x^2 + y^2 \] Given that the height \( y \) is constant at 90 ft, we know that when the kite is 150 ft away from the child, \( x \) can be calculated as: \[ s = \sqrt{x^2 + y^2} = \sqrt{x^2 + 90^2} \] Here, we know the total distance from the child to the kite is 150 ft, so: \[ x^2 + 90^2 = 150^2 \] Calculating this gives: \[ x^2 + 8100 = 22500 \implies x^2 = 14400 \implies x = 120 \text{ ft} \] Now, we differentiate both sides with respect to time \( t \): \[ 2s \frac{ds}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt} \] Since the height \( y = 90 \text{ ft} \) is constant, \( \frac{dy}{dt} = 0 \). Given \( \frac{dx}{dt} = 30 \text{ ft/sec} \), we substitute: \[ 2s \frac{ds}{dt} = 2x \frac{dx}{dt} \] Substituting \( x = 120 \) ft to find \( s \): \[ s = \sqrt{120^2 + 90^2} = \sqrt{14400 + 8100} = \sqrt{22500} = 150 \text{ ft} \] Now we have: \[ 2(150) \frac{ds}{dt} = 2(120)(30) \] This simplifies to: \[ 300 \frac{ds}{dt} = 7200 \] Dividing both sides by 300: \[ \frac{ds}{dt} = \frac{7200}{300} = 24 \text{ ft/sec} \] Thus, the child must let out the string at a rate of \( 24 \text{ ft/sec} \) when the kite is 150 ft away from the child.